Question:easy

Consider that for a supervised learning task, the objective function being minimized is \(f_w(x) = wx\), where \(x \in \mathbb{R}\) is the input and \(w \in \mathbb{R}\) is the parameter. Stochastic Gradient Descent with a learning rate of 0.10 is used for parameter updates.

Suppose that at the end of iteration \(i\), the value of \(w\) becomes 10.00.

Let \(x = 10.00\) be the input for iteration \((i+1)\).

The value of \(w\) at the end of iteration \((i+1)\) is __________. (Rounded off to two decimal places)

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The gradient of wx with respect to w is just x, so the update rule reduces to w minus the learning rate times x.
Updated On: Jul 22, 2026
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Correct Answer: 9

Solution and Explanation

This question is testing the basic SGD update formula, applied to a toy objective that happens to have a very simple gradient.

The general one-parameter SGD step is

\[ w \leftarrow w - \eta \, g \]

where $g$ is the gradient of the objective with respect to $w$, evaluated at the current data point, and $\eta$ is the learning rate.

Here the objective is $f_w(x) = wx$. Treating $x$ as a constant during differentiation with respect to $w$ gives

\[ g = \frac{\partial}{\partial w}(wx) = x \]

So the gradient at any step is simply equal to whatever input $x$ was fed in that step, nothing more elaborate.

Now plug in the numbers given for iteration $(i+1)$: the parameter going into this step is $w = 10.00$ (carried over from the end of iteration $i$), the input is $x = 10.00$, so $g = 10.00$, and $\eta = 0.10$.

The update becomes

\[ w_{new} = 10.00 - (0.10)(10.00) = 10.00 - 1.00 = 9.00 \]

So after this single SGD step, $w$ drops from $10.00$ to $9.00$.

\[ \boxed{w = 9.00} \]
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